Question

Block $A$ of mass $m$ and block $B$ of mass $2 m$ are placed on a fixed triangular wedge by means of a massless, inextensible string and a frictionless pulley as shown in figure. The wedge is inclined at $45^{\circ}$ to the horizontal on both the sides. If the coefficient of friction between the block $A$ and the wedge is $2 / 3$ and that between the block $B$ and the wedge is $1 / 3$ and both the blocks $A$ and $B$ are released from rest, the acceleration of $A$ will be

• A. $- 1\;ms^{-2}$
• B. $- 1.2\;ms^{-2}$
• C. $- 0.2\;ms^{-2}$
• D. zero

Answer 1

Answer: (D)

zero

Answer 2

Total maximum friction force, $f_{t}=f_{A}+f_{B}$
$\mu_{A} m g \cos 45^{\circ}+\mu_{B} 2 m g \cos 45^{\circ}$
or, $f_{t}=m g\left(\frac{2}{3} \times \frac{1}{\sqrt{2}}+\frac{1}{3} \times 2 \times \frac{1}{\sqrt{2}}\right)$
$=\frac{4}{3 \sqrt{2}} mg$
The net pulling force, $F =2 mg \sin 45^{\circ}- mg \sin 45^{\circ}$
$=m g \sin 45^{\circ}$
$=\frac{m g}{\sqrt{2}}$
$\because F < f_{t}$
$\therefore$ The blocks will not move.
So, acceleration of block $A$ is Zero.